Asymptotic normalization coefficients and radiative widths

Abstract

Asymptotic normalization coefficent (ANC) is important quantity in calculation of the radiative width amplitudes providing limits on the radiative width. Here we present some examples showing connection between the ANC and radiative width. In particular, the radiative width of the E1Phys. A 217, 29 (1973) ] is (1.2 ± 0.2) × 10 −2 eV. Meantime the ANC for the first excited state in 17 F puts a low limit on the radiative width, which is (3.4 ± 0.50) × 10 −2 eV. Such a strong disagreement between the measured radiative width and the low limit imposed by the ANC calls for a new measurement of this radiative widh. Other examples are also considered.

Neutron halos in the excited states of B12

“…Application of the ANC theory [19][20][21][22][23] to peripheral reactions on light nuclei is especially adequate for determining the radii of halos [17,24]. In actual practice, the ANC theory is widely used to obtain spectroscopic and astrophysical information [25][26][27][28][29].…”

Internal and external radiative widths in the combined R -matrix and potential-model formalism

“…Therefore, with precise knowledge of these quantities, the channel radiative width amplitude can be calculated quite accurately. The channel radiative width amplitude is a complex quantity and its imaginary part puts a lower limit on the radiative width [6].…”

Extrapolation of scattering data to the negative-energy region

“…Thus the ANC method can be employed as an indirect technique in nuclear astrophysics. The ANCs can be used in evaluating the radiative width of a resonance, decaying to a bound state [6]. An instructive example of using ANCs in nuclear structure studies is the application of the ANC to determine the radii of halo nuclei [7].…”

“…This system involves one bound state corresponding to the ground state of the 6 Li nucleus. An accurate bound-state information is required, e.g.…”

“…for modeling astrophysical α + d → 6 Li + γ radiative capture reaction [19][20][21][22]. 6 Li→αd / √ 2κ = 2.29 of 6 Li in the d + α channel obtained by solving the Faddeev equations for 6 Li without the Coulomb interaction [9].…”